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sin[theta+alpha]/sin[theta+beta]=whole square root over sin2a/sin2b. then the value in tan^2 theta

sohan teja , 6 Years ago
Grade 10
anser 1 Answers
Aditya Gupta

Last Activity: 6 Years ago

cross multiply, and rearrange, you getSin2a/Sin2b= (1-cos2(x+a))/(1-cos2(x+b)) (using 1-2sin^2y=cos2y)Let theta=x and alpha is a  while beta is bsquare both the sides, you get 

sin2a-sin2b=sin2acos2(x+b)-sin2bcos2(x+a)
we now multiply both sides by 2 and apply the standard formulae, to obtain
4sin(a-b)cos(a+b)=sin2(a-x-b)+sin2(x+a-b) [note that here we have used the formulae of sinx-siny, and 2sinxcosy]
this finally yields cos2x=cos(a+b)/cos(a-b)
now using tan^2x=(1-cos2x)/(1+cos2x), we get tan^2x=[cos(a-b)-cos(a+b)]/[cos(a-b)+cos(a+b)]
or tan^2x=tana*tanb
but theta=x
so  tan^2theta=tana*tanb

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